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Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement

Workshop on. Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement. Barisal 21 July 2008 Organized by BSRM Steels Limited. Design Theory on Building Design and Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement.

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Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement

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  1. Workshop on Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement Barisal 21 July 2008 Organized by BSRM Steels Limited

  2. Design Theory on Building Design and Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement Dr. S.M. Atiqul Islam Associate Professor Department of Civil Engineering Dhaka University of Engineering and Technology (DUET) Gazipur-1700 Tel: 8932939 Mobile: 01711389584

  3. Three aims of Civil Engineering design Safety and durability of the structure Beauty of the structure Economy for constructions The responsibility of a design engineer is to fulfill these aims  use of appropriate and quality materials Use of modern technology with the advancement of technology Proper design and method of construction Proper maintenance|

  4. Materials used in Civil Engineering RCC Construction Reinforced concrete is concrete in which reinforcement bars ("rebar") have been incorporated to strengthen the material Reinforcement as Ms. bar Cement Brick Stone Lime, etc. Proper design and appropriate use of these materials ensures the safety and durability of the structure For the economic consideration the use materials should be such that ensure the durability and safety within the affordability

  5. Structural Designconsists the following works  Calculation and analysis of loads.  Realize of effects due to loads.  Findings way to prevent them by selecting and providing proper materials in proper places. In order to understand the structural design it is necessary to understand the failure mechanism of a structure

  6. Flexural failure mechanism of RCC Structure

  7. 40 (300) Grade : Yield Strength = 40 ksi 300 MPa 60 (420) Grade: Yield Strength = 60 ksi 420 MPa 500 (72.5) Grade : Yield Strength = 72.5 ksi or 500 MPa

  8. Typical reinforced concrete beam in walk-though within transparent reinforced concrete

  9. Structural Design • It is required for a Structural Engineer to design the structure to satisfy four major criteria • Appropriateness- The arrangement of spaces, spans, ceiling heights, access and traffic flow must competent to intended use. • Economy- The overall cost of the structure should not exceed the Client Budget • Structural adequacy- A structure must be sufficiently strong to support safety, without collapse, all anticipated loadings and a structure must not deflect, tilt, vibrate, or crack in a manner that impairs its usefulness • Maintainability- A structure should be designed to require a minimum of maintenance and/or to be able to be maintain in a simple fashion

  10. Structural Design • The structural design process is a sequential and iterative decision making process consists of three major phases • Definition of the client’s need and priorities • Development of concept of project • Design of individual system

  11. Design Methods • Usual design methods are based on limit states of the structural elements. Limit states design is a design process that involve the followings • Identification of all potential modes of failure • Determination of acceptable levels of safety against occurrence of each limit state • Consideration by the designer of the significant limit states

  12. Design Methods Generally there are two types of design method to select the concrete and steel reinforcement. First design concept is Working Stress Design and second one is Ultimate Strength Design.

  13. c T= As fs s kd d nAs b In Working Stress Design, members are designed so that stresses in the steel and concrete resulting from normal service loads are within specified limits. These limits, known as allowable stresses, are only fractions of the failure stresses of the materials. Concrete responds reasonably elastically up to compression stresses not exceeding about half its strength, while steel remains elastic practically up to the yield stress. Hence members may be designed on an elastic basis as long as the stresses under service loads remain below these limits.

  14. Column Beam Design • Load • Dead Load: Self wt and furniture, equipment, etc • Live load • Lateral Load; Wind, Earthquake • Impact load (if any)

  15. Lateral Load Earthquake Wind/ Earthquake

  16. From Load Analysis by software or manual Moment and shear can be obtained

  17. Flexural Moment M= 0.5 fc kjbd2 M=Asfsjd

  18. 0.85 f ’c b u = 1c a c d T= As fy y Ultimate Strength Design reinforced concrete structures at loads close to and at failure, one or both of the materials, concrete and steel, are invariably in their nonlinear inelastic range. That is, concrete in a structural member reaches its maximum strength and subsequent fracture at stresses and strains far beyond the initial elastic range in which stresses and strains are fairly proportional. Mn = As fy (d-a/2) where, a= As fy/0.85 f’cb b = 0.851 f’c/fy (87,000/87,000+fy)

  19. The ACI code (2002 and 2005 version) establishes the value of the required strength called U not less than U = 1.2 D + 1.6 L Where D is effect of dead load and L is effect of live load. Other adjustment factors are provided when design conditions involve consideration of the effect of wind, earthquake, differential settlement, creep, shrinkage, temperature change. Design strength of structure is determined by the application of assumptions and requirements given in the code and is further modified by the use of a strength reduction factor  as follows:  = 0.9 for flexure, axial tension, and flexure + tension  = 0.7 for columns with spirals  = 0.65 for columns with ties  = 0.75 for shear and torsion  = 0.65 for compressive bearing

  20. Mu =  Mn and Vu =  Vn Where Mu and Vu is external factored moment and shear forces Mn and Vn nominal ultimate moment and shear capacity of the member Factor of safety for the various values of  and L/D ratio are shown in the following table

  21.  What ever is the design method and what ever is the element for design, strength of concrete and steel are the main factors • If the strength is higher the required amount materials will be less • High strength steel will require less amount of steel • For 60 grade steel required amount is 2/3 of 40 grade • For 500 grade required amount is 20% less than 60 grade

  22. 40 Grade 40 Grade 60 Grade 60 Grade 9- 16 mm 6- 16 mm 6-16 mm 4-16 mm

  23. 40 Grade 40 Grade 60 Grade 60 Grade 6- 16 mm 8- 16 mm 4-16 mm 4-16 mm

  24. USD Design Slab ACI code provision for minimum slab thickness (9.5.2.1 table 9.5 (a)) 9.5 (a) Minimum thickness of nonprestressed beams or one way slabs unless defection are calculated

  25. USD Design Slab 9.5 (c) Minimum thickness of slabs without interior beams

  26. USD Design Slab ACI code minimum reinforcement 7.12.2.1 (b) slab where 60 grade deformed bar or welded wire reinforcement are used ….0.0018 (c) slab where reinforcement with yield stress exceeding 60,000 psi measured at a yield strain of 0.35 percent is used …..7.12.2.2 Shrinkage and temperature reinforcement shall be spaced not farther apart than 5 times the slab thickness, nor farther apart than 18 in.

  27. USD Design Slab

  28. USD Design Slab One way slab

  29. USD Design Column Factored Moment = 30 kip-ft Pu = 330 kip Size = 12” x 12” Designed by 420 grade steel required reinforcement = 4 # 8 bars Designed by 500 grade steel required reinforcement = 4 # 7 bars Reinforcement ratio 0f 420 grade to 500 grade = 2.4/3.16 = 0.76 = 76% Savings in steel = 26% 4 # 8 (grade 420) 4 #7 (grade 500)

  30. USD Design Beam For a beam it is known that, Mn = As * fy * (d-a/2) Beam with Grade 420 (60) Mn = As60 * fy60 * (d-a/2) Beam with Grade 500 Mn = As500 * fy500 * (d-a/2) As75 = (fy420/fy500) * As420 As500 = 0.82 * As420 Savings in steel = 20%

  31. Cost Saving in Steel by Using 500-Graded Rebar in Place of Ordinary Steel As Reinforcement

  32. In reinforced concrete a long trend is evident towards the use of higher strength materials, both steel and concrete. Reinforcing bars with 40 ksi (40 grade) yield stress, almost standard 20 years ago, has largely been replaced by 60 ksi (60 grade) yield stress. Now a days the demand of high strength steel is increasing. The use of steel 70-100 ksi is increasing day by day. This is because they are more economical and their use tends to reduce congestion of steel in the forms.

  33. Table 1 lists presently available reinforcing steel, their grade designation, ASTM specifications. The yield strength of 500 grade is 500 MPa or 72.5 ksi.

  34. The marking system of rebars to meet ASTM specifications. 60 Grade or higher 40 Grade or higher

  35. Stress-strain diagram for differernt graded steel

  36. The major advantages of using 500 graded steel over any other steel can be summarized: • As the steel requirements is lower for 500 graded steel, it is very economical; • The steel requirement is lower, the steel congestion relief in the form is lower in the form as shown in Figure 9, Therefore the handling of steel in form is easier • As the steel congestion in the form is lower, better concrete placement can be achieved and medium to large course aggregate can be used and higher strength of concrete can be achieved (a) (b)

  37. Better bonding is achieved for 500 graded steel • As higher yield strength is achieved in 500 Graded Steel than specified, then it is more safer to use 500 Graded steel in reinforcement. • For heavy construction, the code recommendation is to use 420 (60) Grade or higher grade steel. (a) (b)

  38. Example of lower steel area requirement for 500 graded steel The following table shows the steel requirement for 500, 420 (60) and 300 (40) graded steel and preferable bar steel. The area of the steel can be calculated by the following: Area required by 500 graded steel = 420/500*Area required by 420 grade = 300/500*Area required by 300 grade Cost savings by using 500 graded steel considering 1 ton of steel of 300 grade

  39. Example of cost savings by using 500 graded steel

  40. Example of cost savings by using 500 graded steel Cost saving by using 500 graded over 300 grade steel = (70,200-40,260)* 100/70,200= 42.6% Cost saving by using 500 graded over 420 grade steel = (49,045-40,260)* 100/49,045= 18.0% Cost saving by using 420 graded over 300 grade steel = (70,200-49,045)* 100/70,200= 30.1% However, using of 500 grade steel will increase the 10-15% of concrete volume from 300 grade and 5-10% concrete volume from 420 grade, which will reduce the savings and fall to around 12-15% over 420 grade and 30-35% over 300 grade steel

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